# Thread: Evaluate the derivative using properties of logarithms where needed.

1. ## Evaluate the derivative using properties of logarithms where needed.

As some added context, I am taking Calculus: Early Transcendental Functions and we are studying "The Natural Logarithm as an Integral"

The problem presented is as follows:

d/dx [ln (x5 sin x cos x)]

I've read through the material and even looked at some of the odd numbered problems that have answers but I just can't seem to get started here. Any advice on how to approach this?

2. ## Re: Evaluate the derivative using properties of logarithms where needed.

I think I would simplify the application of the chain rule a bit by first writing the expression as:

$\displaystyle \frac{d}{dx}\left(\ln(x^5\sin(2x))-\ln(2) \right)$

Now, the derivative of the constant $\displaystyle \ln(2)$ is zero, so we are left with:

$\displaystyle \frac{d}{dx}\left(\ln(x^5\sin(2x)) \right)$

See if you can now apply the rule:

$\displaystyle \frac{d}{dx}\left(\ln(u(x)) \right)=\frac{1}{u(x)}\cdot\frac{du}{dx}$

3. ## Re: Evaluate the derivative using properties of logarithms where needed.

Originally Posted by JDS
As some added context, I am taking Calculus: Early Transcendental Functions and we are studying "The Natural Logarithm as an Integral"

The problem presented is as follows:

d/dx [ln (x5 sin x cos x)]

I've read through the material and even looked at some of the odd numbered problems that have answers but I just can't seem to get started here. Any advice on how to approach this?
Following Mark's example of simplifying the logarithm, you should simplify even further before trying to take the derivative. Write it as \displaystyle \displaystyle \begin{align*} y = 5\ln{(x)} + \ln{\left[ \sin{(x)} \right]} + \ln{\left[ \cos{(x)} \right]} \end{align*} and then apply the much simpler chain rules to each term.

4. ## Re: Evaluate the derivative using properties of logarithms where needed.

Thanks for the quick reply! Here is what I came up with and hopefully I have not confused myself!

= (1/x5 sin2x) * x5

= 1/sin2x

5. ## Re: Evaluate the derivative using properties of logarithms where needed.

Originally Posted by Prove It
Following Mark's example of simplifying the logarithm, you should simplify even further before trying to take the derivative. Write it as \displaystyle \displaystyle \begin{align*} y = 5\ln{(x)} + \ln{\left[ \sin{(x)} \right]} + \ln{\left[ \cos{(x)} \right]} \end{align*} and then apply the much simpler chain rules to each term.
By following that, I get....

Y = 5 ln (x) + ln[sin(x)] + ln[cos(x)]

5 ln (x) + ln[cos(x)] + ln[-sin(x)]

grrr, and I am defintely sure I am confused now! LOL!

6. ## Re: Evaluate the derivative using properties of logarithms where needed.

Let's take a look at both approaches:

My approach:

$\displaystyle \frac{d}{dx}(\ln(x^5\sin(2x)))=\frac{x^5(2\cos(2x) )+5x^4\sin(2x))}{x^5\sin(2x)}=2\cot(2x)+5x^{-1}$

Prove It's approach:

$\displaystyle \frac{d}{dx}(5\ln(x)+\ln(\sin(x))+\ln(\cos(x)))=5x ^{-1}+\frac{\cos(x)}{\sin(x)}-\frac{\sin(x)}{\cos(x)}=$

$\displaystyle 2\cot(2x)+5x^{-1}$

My approach has a more difficult application of the chain rule, but no need to apply double-angle identities (at least if you wish to combine the two trig. terms).